• Oct 19th 2011, 12:35 PM
Niall101
The answer I get is the intersection of all 3 is <= -5!

But it should be a positive integer in my mind.

I have worked this out about 10 times now and I get -5 every time!

Any help mch appreciated!
• Oct 19th 2011, 01:46 PM
ZeroDivisor
The problem does not make much sense to me.
Suppose the first 15 out of 25 attend session one.
Suppose the 10 other plus 8 of the first 15 attend session two (so that 7 of the ones that attended the first session do not attend the second).
Then you could have that the 10 other again and two of the 7 attend the third. That way, no one attends all three.
• Oct 19th 2011, 07:33 PM
Prove It
Quote:

Originally Posted by Niall101
The answer I get is the intersection of all 3 is <= -5!

But it should be a positive integer in my mind.

I have worked this out about 10 times now and I get -5 every time!

Any help mch appreciated!

$\displaystyle n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap C) - n(B \cap C) - n(A \cap B \cap C)$
• Oct 20th 2011, 02:00 AM
emakarov
Quote:

Originally Posted by Prove It
$\displaystyle n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap C) - n(B \cap C) - n(A \cap B \cap C)$

How does this help here? I used some lower bounds on pairwise intersections, but I got that $n(A \cup B \cup C)\ge-5$, which is not much.
• Oct 20th 2011, 04:41 AM
Niall101
hi i used this too and got -5 also.
• Oct 20th 2011, 06:45 AM
Deveno
the trouble is, the estimates on the two-set intersections are all done independently, and they're not truly independent of each other. for example, it's easy to establish that |A∩B| must be at least 8, but if we assume it IS 8, we get "higher estimates" for A∩C and B∩C.

however, the minimum, as the problem is stated is indeed 0.

asuume A∩B = 8. then 7 people attended the 1st but not the 2nd session, and 10 people attended the 2nd but not the 1st session.

all of the people who attended the 3rd session (12 people), must then have attended the 1st and 3rd only, the 2nd and 3rd only, or all 3.

no more than 7 of these 12 could have attended just the 1st and 3rd, because only 7 attended the 1st but not the 2nd.

the remaining 5 could have attended just the 2nd and 3rd, giving the following breakdown:

8 people attended the 1st and 2nd only
7 people attended the 1st and 3rd only
5 people attended the 2nd and 3rd only
5 people attended only the 2nd.

summing, we have: 25 = |AUBUC| = |A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C| = 15+18+12-8-7-5 = 45-20 = 25, so we see it is possible that no one attended all 3.
• Oct 20th 2011, 10:08 AM
Niall101